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Updated on: 20 Sep 2015, 06:44

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Difficulty:

25% (medium)

Question Stats:

83%(01:45) correct 17%(01:57) wrong based on 327 sessions

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At 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph. If, at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph, at what time will he catch up to Peter?

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10 Aug 2009, 00:29

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crejoc wrote:

At 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph. If, at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph, at what time will he catch up to Peter? 6:00 p.m.6:30 p.m.8:00 p.m.8:30 p.m.10:00 p.m

since Peter had early start of 4 hrs, in this 4 hrs he covered 40miles. after that John starting in the same path.the differences in speed is (15-10)mph = 5mph, in order to catch up Peter, john has to cover the above 40miles.=40/5 = 8 hrs i.e 8 hrs from 2:00 pm is equal to 10:00p m

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21 Sep 2015, 05:19

1

crejoc wrote:

At 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph. If, at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph, at what time will he catch up to Peter?

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10 Nov 2016, 23:05

crejoc wrote:

At 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph. If, at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph, at what time will he catch up to Peter?

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14 Nov 2016, 07:28

1

crejoc wrote:

At 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph. If, at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph, at what time will he catch up to Peter?

A. 6:00 p.m.B. 6:30 p.m.C. 8:00 p.m.D. 8:30 p.m.E. 10:00 p.m

We can classify this problem as a “catch-up” rate problem, for which we use the formula:

distance of Peter = distance of John

We are given that at 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph, and that at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph.

Since Peter started 4 hours before John, we can let Peter’s time = t + 4 hours, and John’s time = t.

Since distance = rate x time, we can calculate each person’s distance in terms of t.

Peter’s distance = 10(t + 4) = 10t + 40

John’s distance = 15t

We can equate the two distances and determine t.

10t + 40 = 15t

40 = 5t

t = 8 hours

Since John left at 2 p.m. and caught up to Peter 8 hours later, he caught up with Peter at 10 p.m.